ACM平衡的括号

最新推荐文章于 2022-02-19 09:49:55 发布

转载最新推荐文章于 2022-02-19 09:49:55 发布 · 60 阅读

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原文链接：http://www.cnblogs.com/xinxiangqing/p/4669649.html

Description

You are given a string consisting of parentheses () and []. A string of this type is said to be correct:

(a)

if it is the empty string

(b)

if A and B are correct, AB is correct,

(c)

if A is correct, (A ) and [A ] is correct.

Write a program that takes a sequence of strings of this type and check their correctness. Your program can assume that the maximum string length is 128.

Input

The file contains a positive integer n and a sequence of n strings of parentheses () and [], one string a line.

Output

A sequence of Yes or No on the output file.

Sample Input

3
([])
(([()])))
([()[]()])()

Sample Output

#include <iostream>
#include <cstdio>
#include <string.h>
#include <stack>
using namespace std;
int main()
{
    char s[150];
    int n;
    cin>>n;
    getchar();
    while(n--)
    {
        gets(s);
        stack<char>v;
        if(strlen(s)==0)
        {
            cout<<"Yes"<<endl;
            continue;
        }
        for(int i=0;i<strlen(s);i++)
        {
            if(s[i]=='('||s[i]=='['||(v.empty()))
                v.push(s[i]);
            else
            {
                if((s[i]==')'&&v.top()=='(')||(s[i]==']'&&v.top()=='['))
                    v.pop();
                else
                    v.push(s[i]);
            }
        }
        if(v.size())
            cout<<"No"<<endl;
        else
            cout<<"Yes"<<endl;
                
    }
    return 0;
}

Yes
No
Yes

解题思路：这个问题推荐用栈来解决，我们小建立一个字符数组用来存储从外面输入进来的字符串。对于'( '和' [ '的字符以及当栈是空的时候我们就往栈里面存放字符，否则如果此时要存入的字符是')'（且栈的顶部的元素是'('）或‘ ]‘（此时的栈的顶部的元素是'['）我们就把栈的顶部的元素拿出栈，否责就继续把字符存入栈。最后我们判断栈是空的还是满的，我们就能确定我们输入的字符串是不是平衡的了；如果栈是空的，我们就输出Yes，表示这个字符串是平衡的，否则输出No。在这个题目中我们还应该注意的是当输入的字符串是空的时候，这时的字符串也是平衡的。

转载于:https://www.cnblogs.com/xinxiangqing/p/4669649.html